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Viefleur [7K]
2 years ago
14

How can you find 8 people can share 7 tickets equally

Mathematics
2 answers:
9966 [12]2 years ago
5 0

Step-by-step explanation:

8/7 because 7*8=7/1*8/1= and we will take the reciprical which is 7/1*1/8=7/8

densk [106]2 years ago
3 0

Answer:

Easy Kick one person out

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Kevin is 6 years older than Timothy. 4 years ago, Kevin was twice as old as Timothy. Find their present ages.
Novosadov [1.4K]

Step-by-step explanation:

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6 0
2 years ago
If p is inversely proportional to q and p=-8 then q=2, find p when q=-4
Sunny_sXe [5.5K]

Answer:

16

Step-by-step explanation:

5 0
3 years ago
You separate 42 bulbs of garlic into two groups:one for planting and one for cooking. You will plant 3 bulbs for every 4 bulbs t
alexira [117]

Answer:

144 cloves

Step-by-step explanation:

You separate 42 bulbs of garlic into two groups:one for planting and one for cooking.

We have a ratio of planting to cooking

Planting : Cooking

You will plant 3 bulbs for every 4 bulbs that you will use for cooking.

Hence: 3 : 4

The total of their proportion = 3 + 4 = 7

Hence:

The number of garlic you plant =

3/7 × 42 = 18 bulbs of garlic

The number of garlic you use to cook =

4/7 × 42 = 24 bulbs of garlic

We are also told that:

Each bulb has about 8 cloves. How many cloves of garlic will you plant?

From the calculation above , we know the number of bulbs of garlic been planted

= 18 bulbs.

Hence:.

1 bulb = 8 cloves

18 bulbs = x

x = 8 cloves × 18 bulbs

x = 144 cloves

Therefore, you will plant 144 cloves of garlic

6 0
3 years ago
Explain how to write a quadratic equation given the following three points on the graph (5,31) (3,11) (0,11)
masha68 [24]

Given:

The graph of a quadratic function passes through the points (5,31) (3,11) (0,11).

To find:

The equation of the quadratic function.

Solution:

A quadratic function is defined as:

y=ax^2+bx+c            ...(i)

It is passes through the point (0,11). So, substitute x=0,y=11 in (i).

11=a(0)^2+b(0)+c

11=c

Putting c=11 in (i), we get

y=ax^2+bx+11               ...(ii)

The quadratic function passes through the point (5,31). So, substitute x=5,y=31 in (ii).

31=a(5)^2+b(5)+11

31-11=a(25)+5b

20=25a+5b

Divide both sides by 5.

4=5a+b                  ...(iii)

The quadratic function passes through the point (3,11). So, substitute x=3,y=11 in (ii).

11=a(3)^2+b(3)+11

11-11=a(9)+3b

0=9a+3b

Divide both sides by 3.

0=3a+b                 ...(iv)

Subtracting (iv) from (iii), we get

4-0=5a+b-3a-b

4=2a

\dfrac{4}{2}=a

2=a

Putting a=2 in (iv), we get

0=3(2)+b

0=6+b

-6=b

Putting a=2,b=-6 in (ii), we get

y=(2)x^2+(-6)x+11

y=2x^2-6x+11

Therefore, the required quadratic equation is y=2x^2-6x+11.

7 0
2 years ago
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